If p is a natural number and p! ends with y trailing zeros

I solved it this way:

Since the p! has trailing zeros, Let's assume p=10

10! will have 2 trailing 0s (by the method provided by Bunuel)
p = 10 y = 2
5p! i.e 50! will have 12 trailing 0s = 10 + 2 = p + y

for p = 10 trailing 0's =
10/5 = 2

for 5p,
50/5 = 10, 50/25 = 2 hence total is 12
Similarly,
for p = 20 trailing 0's = 20/5 = 4
for 5p, trailing 0's = 50/5 = 10, 50/25 = 2 hence total is 10 + 2 = 12
thus we observe,
number of 0's = p + y
example p = 20, y = 4 giving 24.

Hence A.

suppose p=6
I am taking 6 because it will have 5 & 2 both to become trailing zeroes.

now 6! has 1 trailing zero.

=6/5 = 1 =y

so 5p = 5*6 = 30

new number is 30!.
Trailing zeroes in (5p)! = 30/5 + 30/25 = 6+1 = 7
now here we get 7=6+1=p+y

which is our answer...opt A

Hi Bunuel,

I'm trying to understand the explanation here and am unable to understand how we got the last step:

now, # of trailing zeros for (5p)! will be \frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...=p+(\frac{p}{5}+\frac{p}{5^2}+...)=p+y.

I understand that the # of trailing zeros for (5p)! will be \frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+...

But how is this equal to p+(\frac{p}{5}+\frac{p}{5^2}+...)=p+y?

Infact when I solve \frac{5p}{5}+\frac{5p}{5^2}+\frac{5p}{5^3}+... I factor out 5 and get 5(\frac{p}{5}+\frac{p}{5^2}+\frac{p}{5^3}+...)= 5y

This might be a silly question and I'm definitely missing something out here... but can't figure out where I'm going wrong.

Kindly help me out.

Thanks.

P.S: This is the first time I'm posting on this forum... Not sure If I've done it right. Please let me know if anything needs to be changed.

...Or use smart numbers

p= 5! we have 1 trailing zero

5*5! = 25! we have 6 trailing zeroes

Since p = 5 and 'y' = 1

The only answer choice that will satisfy is A

Hope it helps
Cheers!
J

Chandni170 . you have a problem with your last denominator....
- assume the last denominator for the P! division is 5^n .
- then the last denominator of (5P)! is not 5^n but 5^(n+1) .
now if you factor out 5 you will end up with 5 *(y+P/5^(n+1)) and not 5*y . Hope it helps...

Let p = 1. p! = 1! = 1, which means y= 0

(5p)! = 5! = 120, trailing zeros = 1

1 = 1 + 0 = p + y

Answer (A).

p=5
p!= 120 has one zero

So, p=5 then y=1

Then

5p! = 25!

Counting zeros in 25!

5*2 =10
10
15*6 = 90
20
24*5=120
25*4=100

So 25!= 5p! has six zeros

Number of zeros in 5p! = 6 = 5 + 1 = p +y

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